Given that $n$ is a natural number, the expression $(n+1)(n+2)\cdots(n+10)$ represents the product of 10 consecutive integers. The question asks which number this product cannot be.

The product of 10 consecutive natural numbers is divisible by several small prime numbers. Specifically:

• One of these numbers is divisible by 2.
• One is divisible by 3.
• One is divisible by 5.
• One is divisible by 7.

Thus, the product will always include factors of 2, 3, 5, and 7. Therefore, the product of these numbers will be divisible by:

• $2 \times 3 \times 5 \times 7 = 210$

This means that $(n+1)(n+2)\cdots(n+10)$ will always be divisible by 210.

Now, let's consider a specific number that this product cannot be. Since the product will always be divisible by 210, it cannot be a prime number greater than 210, because prime numbers have no divisors other than 1 and themselves. Therefore, the product of 10 consecutive integers can never be a prime number greater than 210.

Would you like more details or have any questions?

Here are five follow-up questions for exploration:

1. Why is the product of 10 consecutive integers always divisible by 210?
2. Can you generalize this reasoning to other products of consecutive integers, such as 5 or 15 integers?
3. What prime number properties can be used to predict divisibility in such problems?
4. How can we express the prime factorization of a product like $(n+1)(n+2)\cdots(n+10)$?
5. Could this product ever result in a perfect square? Why or why not?

Tip: Always check the divisibility properties of consecutive numbers by examining their prime factors.